Question

Two particles of equal mass $m$ have respective initial velocities $u\hat{i}$ and $u\left(\frac{\hat{i}+\hat{j}}{2}\right)$They collide completely inelastically. The energy lost in the process is

• A. $\frac{1}{8}m{u}^2$
• B. $\frac{3}{4}m{u}^2$
• C. $\sqrt{\frac{2}{3}}m{u}^2$
• D. $\frac{1}{3}m{u}^2$

Answer 1

The correct option is A $\frac{1}{8}m{u}^2$

Step 1: Find the common velocity of the particles.
The given collision is completely inelastic and both the particles will stick together and move with common velocity.
From conservation of linear momentum.
$mu\hat{i}+mu\left(\frac{\hat{i}+\hat{j}}{2}\right)=(m+m)v$
$v=\frac{3}{4}u\hat{i}+\frac{u}{4}\hat{j}$
$\left|v\right|=\frac{u}{4}\sqrt{10}$

Step 2: Find the final and kinetic energy of the system
Final kinetic energy $=\frac{1}{2}2m{\left(\frac{u}{4}\sqrt{1}0\right)}^2=\frac{5}{8}m{u}^2$
Initial kinetic energy$=\frac{1}{2}m{u}^2+\frac{1}{2}m{\left(\frac{u}{\sqrt{2}}\right)}^2=\frac{6}{8}m{u}^2$

Step 3: Calculate the energy lost in the process.
Loss in $K.E=k_i-k_f=\frac{1}{8}m{u}^2$